If a weak acid is 25% deprotonated at ph 4, what is the pka?if a weak acid is 25% deprotonated at ph 4, what is the pka?

Answer:

86g of [tex]Al_{2}O_{3}[/tex]

Explanation:

1. The balanced chemical equation is:

[tex]_{4}Al+_{3}O_{2}=_{2}Al_{2}O_{3}[/tex]

2. Use the stoichiometry of the reaction to calculate how many moles of aluminum oxide can be formed.

If the oxygen is in excess it means that the aluminum is the limiting reagent and the calculations must be made using the mass of aluminum.

[tex]45.3gAl*\frac{1molAl}{27gAl}*\frac{2molesAl_{2}O_{3}}{4molesAl}=0.84molesAl_{2}O_{3}[/tex]

3. Use the molar mass of the [tex]Al_{2}O_{3}[/tex] to calculate how many grams of aluminum oxide can be formed:

[tex]0.84molesAl_{2}O_{3}*\frac{102gAl_{2}O_{3}}{1molAl_{2}O_{3}}=86gAl_{2}O_{3}[/tex]

Final answer:

The empirical formula of phosphorus selenide is represented as P₂Sen, where 'n' is the number of selenium atoms bonded to two phosphorus atoms. Specific stoichiometry is needed for an exact formula.

Explanation:

The empirical formula of a compound represents the simplest whole-number ratio of the elements within that compound. For phosphorus selenide, we need to look at the actual stoichiometry of the compound to determine its empirical formula. Without specific information on the stoichiometry, we cannot give an exact empirical formula. However, some known phosphorus selenides include P₂Se₃ and P₂Se. To generalize, an empirical formula for phosphorus selenide could be P₂Sen, where 'n' represents the number of selenium atoms that bond with two phosphorus atoms.

Answer: Helium

Explanation:

Answer:

There are two P (phosphorous) atoms and five O (oxygen) atoms.

Explanation:

The subscripts of each element represent the number of atoms of  that element in the molecule.

Atom is the smallest unit of matter. An element is made of identical atoms.

A molecule is made up of more than one different element. And there are more than one atoms of each element in the molecule.

The secondary structure of a protein is determined by hydrogen bonds between amino acids in different regions of the polypeptide chain.

The secondary structure of a protein results from hydrogen bonds between amino acids in different regions of the polypeptide chain. This folding can take the form of an alpha-helix or a beta-pleated sheet. Hydrogen bonds form between the oxygen atom in the carbonyl group of one amino acid and the amino acid that is four amino acids farther along the chain.

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The empirical formula of the new oxide is MoO₃.

1) Find the number of moles of Mo₂O₃: n= mass/Mwt=(18.75)/(239.9)=0.0782 mol of Mo₂O₃ 2)

Find the number of moles of Mo in Mo₂O₃: 1 mol Mo2O3 --> 2 moles of Mo

0.0782 mol of Mo₂O₃ --> X moles of Mo X = (2×0.0782)/1

=0.156 mol of Mo

In the first oxide (it would be the same for the second oxide “see step 3”) 3) The mass of Mo in the second oxide doesn’t change (because the question said that oxygen has been added “only”, therefore, the mass of Mo in the first oxide = the mass of Mo

in the second oxide: Mass of Mo = n x Mwt of Mo

= 0.156 x 95.95

= 14.97 grams.

4) Find the mass of oxygen in the second oxide: mass of the second oxide = mass of Mo + mass of O

 Mass of O = Mass of the second oxide – Mass of Mo

= 22.50 – 14.97

= 7.53 grams

5) Find the empirical formula by finding their number of moles first, then divide it by the smallest number of moles:

mass of Mo = 14.97 grams --> moles of Mo

= 0.156 mol of Mo

mass of O = 7.53 grams

moles of O = 0.471 mol of O

Divide by the smallest number of moles (by 0.156)

Therefore, the new empirical formula is MoO₃.

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The empirical formula of the new oxide is MoO₄.

To identify the empirical formula of the new oxide, we need to calculate the number of moles of molybdenum and oxygen in the new oxide.

Moles of molybdenum:

Moles of molybdenum = 11.340 g / 95.94 g/mol = 0.118 mol

Moles of oxygen:

Moles of oxygen = (11.340 g - 10.63 g) / 16.00 g/mol = 0.0436 mol

Empirical formula:

The empirical formula of the new oxide is MoO_{x}, where x is the number of moles of oxygen per mole of molybdenum. We can calculate x by dividing the moles of oxygen by the moles of molybdenum:

x = 0.0436 mol / 0.118 mol = 0.370

Rounding to the nearest integer, we get x = 4. Therefore, the empirical formula of the new oxide is MoO₄.

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The relationship between an element's atomic mass and an element's isotopes is that they have the same number of protons in their respective nuclei.

So, from this, we can conclude that the relationship between an element's atomic mass and an element's isotopes is that they have the same number of protons in their respective nuclei.

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Final answer:

To find the number of moles in a given mass of a substance, divide the mass by the molar mass. To find the mass of a given number of moles of a substance, multiply the number of moles by the molar mass.

Explanation:

To find the number of moles in a given mass of a substance, we use the formula:

moles = mass / molar mass

For example:

1. To find the number of moles in 28.0 grams of oxygen:

moles = 28.0 g / 32.00 g/mol (molar mass of oxygen)

moles = 0.875 mol

2. To find the mass of 5.0 moles of iron:

mass = 5.0 mol * 55.85 g/mol (molar mass of iron)

mass = 279.25 g

3. To find the number of moles of argon in 452 g of argon:

moles = 452 g / 39.95 g/mol (molar mass of argon)

moles = 11.31 mol

4. To find the grams in 16.5 mol of hydrogen:

mass = 16.5 mol * 1.01 g/mol (molar mass of hydrogen)

mass = 16.665 g

The Celsius temperature required to change the volume of the gas sample to a specific volume at constant pressure can be found using Charles's Law. In this case, the temperature, volume, and initial conditions are given, allowing us to solve for the unknown temperature. By substituting the values into the formula, we find that the required temperature is 316 K.

A volume change caused by a temperature change at constant pressure means we should use Charles's law. Taking V₁ and T₁ as the initial values, T2 as the temperature at which the volume is unknown and V₂ as the unknown volume, and converting °C into K we have:

T₂ = (T₁ * V₂) / V₁

Substituting the given values:

T₂ = (43 + 273) * (2.26 × 10³) / (1.13 × 10³)

T₂ = 316 K

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To change the gas volume from 1.13×10³ L at 43°C to 2.26×10³ L, the required temperature is 359°C. This is calculated using Charles's Law and converting temperatures between Celsius and Kelvin.

Calculating the Required Temperature Using Charles's Law

To solve this, we will use Charles's Law, which states that the volume of a gas is directly proportional to its temperature (in Kelvin) when pressure and the amount of gas are constant. The formula for Charles's Law is:

V₁ / T₁ = V₂ / T₂

First, we convert the temperatures from Celsius to Kelvin:

T₁ = 43°C + 273 = 316 K

Next, we rearrange Charles's Law to solve for T₂:

T₂ = (V₂ × T₁) / V₁

Substitute the known values into the equation:

T₂ = (2.26×10³ L × 316 K) / 1.13×10³ L

T₂ = 632 K

Finally, convert the temperature from Kelvin back to Celsius:

T₂ = 632 K - 273 = 359°C

Thus, the required temperature t₂ is 359°C.

Helium and Sodium are elements, Water and Salt are compounds, while Sedimentary Rock and Air are mixtures.

To correctly classify each substance, we need to understand what elements, compounds, and mixtures are. An element is a pure substance consisting of only one type of atom. A compound is a substance made up of two or more different elements that are chemically bonded. A mixture is composed of two or more different substances that retain their own properties and can be separated physically.

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Answer:  Flash Point

Explanation:  Volatile substance is the substance which can easily catch fires.

Thus, there must be a point to define that temperature when the volatile substance catches fire given enough source of flame. This point is known as Flash Point.

Thus, Flash point is defined as the lowest temperature at which the flammable or volatile liquid catches fire and convert itself into the ignitable mixture when enough source of ignition is given.

There is a gravitational force pulling the stack of books downward. Gravity is always a factor on earth. Keep that in mind when asked these sort of questions. They are made to confuse you. Take your time and you will do just fine.

The amount of [tex]{{\text{N}}_2}[/tex]  initially taken is [tex]\boxed{{\text{1}}{\text{.5 mol}}}[/tex] and the amount of [tex]{{\text{H}}_2}[/tex]  initially taken is  [tex]\boxed{{\text{2}}{\text{.5 mol}}}[/tex].

Further explanation:

Stoichiometry of a reaction is used to determine the amount of species present in the reaction by the relationship between the reactants and products. It can be used to determine the moles of a chemical species when the moles of other chemical species present in the reaction is given.

Consider the general reaction,

 [tex]{\text{A}}+2{\text{B}}\to3{\text{C}}[/tex]

Here,

A and B are reactants.

C is the product.

One mole of A reacts with two moles of B to produce three moles of C. The stoichiometric ratio between A and B is 1:2, the stoichiometric ratio between A and C is 1:3 and the stoichiometric ratio between B and C is 2:3.

The balanced chemical equation for the formation of [tex]{\text{N}}{{\text{H}}_3}[/tex]  is as follows:

[tex]{{\text{N}}_2}\left(g\right)+3{{\text{H}}_2}\left(g\right)\to2{\text{N}}{{\text{H}}_3}\left(g\right)[/tex]

The balanced chemical equation shows that 1 mole of [tex]{{\text{N}}_2}[/tex]  and 3 moles of [tex]{{\text{H}}_2}[/tex]  reacts to form 2 moles of [tex]{\mathbf{N}}{{\mathbf{H}}_{\mathbf{3}}}[/tex] .

In the question, reaction started with the unknown quantity of reactant [tex]{{\text{N}}_2}[/tex]  and [tex]{{\text{H}}_2}[/tex] and stopped when 1.0 mol ammonia was produced. Also, reactants left in the reaction mixture are 1.0 mol of [tex]{{\text{H}}_2}[/tex]  and 1.0 mol of [tex]{{\text{N}}_2}[/tex] .

According to the balance reaction 2 moles of ammonia is produced by the 3 moles of [tex]{{\text{H}}_2}[/tex] . Thus amount of hydrogen molecule required to produce 1 mole of ammonia is,

[tex]\begin{gathered}{\text{Moles of }}{{\text{H}}_2}=\frac{{3{\text{ mol }}{{\text{H}}_2}}}{{2{\text{ mol N}}{{\text{H}}_3}}}\times1{\text{ mol N}}{{\text{H}}_3}\\=1.5{\text{ mol }}{{\text{H}}_{\text{2}}}\\\end{gathered}[/tex]

According to the balance reaction 2 moles of ammonia is produced by the 1 mole of [tex]{{\text{N}}_2}[/tex] . Thus the amount of [tex]{{\text{N}}_2}[/tex] molecule required to produce 1 mole of ammonia is,

[tex]\begin{gathered}{\text{Moles of }}{{\text{N}}_2}=\frac{{1{\text{ mol }}{{\text{N}}_2}}}{{2{\text{ mol N}}{{\text{H}}_3}}}\times1{\text{ mol N}}{{\text{H}}_3}\\=0.5{\text{ mol }}{{\text{N}}_2}\\\end{gathered}[/tex]

Therefore, the amount of [tex]{{\text{N}}_2}[/tex]  and [tex]{{\text{H}}_2}[/tex]  are consumed until the reaction is stopped is 0.5 moles and 1.5 moles respectively.

Therefore, the amount of [tex]{{\text{N}}_2}[/tex]  initially taken is,

[tex]\begin{aligned}{\text{Amount of }}{{\text{N}}_2}&=\left({1.0 + 0.5}\right){\text{ mol}}\\&=1.5{\text{ mol}}\\\end{aligned}[/tex]

The amount of [tex]{{\text{H}}_2}[/tex]  initially taken is,

[tex]\begin{aligned}{\text{Amount of }}{{\text{H}}_2}&=\left({1.0 + 1.5}\right){\text{ mol}}\\&=2.5{\text{ mol}}\\\end{aligned}[/tex]

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Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Mole concept

Keywords: N2, H2, NH3, 3H2, 2NH3, limiting reagent, nitrogen, hydrogen, ammonia, 0.5 mol, 1.0 mol, 1.5 mol, 2.5 mol, 1.0 mol of NH3.

The number of moles of N2 and H2 that were originally present are;

N2 = 1.5 mol

N2 = 1.5 molH2 = 2.5 mol

This reaction when balanced is;

N2(g) + 3H2(g) = 2NH3(g)

Thus,divide each of the number of moles in the balanced equation by 2 to get;

0.5N2(g) + 1.5H2(g) = NH3(g)

1 mol of N2 and 1 mol of H2.

Thus;

Number of moles of N2 originally present = 1 + 0.5 = 1.5 moles

Number of moles of H2 originally present = 1 + 1.5 = 2.5 moles

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The answer is transition from Y to Z, has a greater energy difference because the red light means higher energy and the blue light means lower energy.

You would be able to find this on the diagram of electromagnetic spectrum.

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Answer:

Answer:

(an open) (30) (greater than 30) ( 40

Explanation:

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